Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Fri 5/27 Learning Objective: To analyze normal distribution Lesson Hw: 11 – 10 WS 1 11 – 10 Algebra II To find normal distribution Discrete Probability Distribution – has a finite number of possible events, or values Continuous Probability Distribution – Events can be any value in an interval of real numbers. Large data set Normal Distribution – data that vary randomly from the mean. Normal Curve Skewed Data – distribution is an asymmetric curve where one end stretches out further than the other end - Data do not vary predictably from the mean - Do not use mean & standard deviation to estimate percentages for skewed data 1. Sketch a normal curve for each of the distribution. Label the x – axis values at one, two, and three standard deviation from the mean. Mean = 15.7 Std Deviation = 2.8 34% 34% 13.5% 2.35% 2.35% 13.5% 7.3 10.1 12.9 15.7 18.5 21.3 24.1 2. Sketch a normal curve for each of the distribution. Label the x – axis values at one, two, and three standard deviation from the mean. Mean = 21.1 Std Deviation = 4.7 4.7 4.7 4.7 34% 34% 4.7 13.5% 2.35% 2.35% 13.5% 7 11.7 16.4 21.1 25.8 30.5 35.2 3. The bar graph gives the weights of a population of female brown bears. The curve shows how the weights are normally distributed about the mean, 115 kg. Approximately what percent of the female brown bears weight between 100 and 129 kg. 23% + 42% + 23% = 88% 4. Approximately what percent of the female brown bears weight less than 120 kg. 23% + 42% + 5% = 70% less than 120 kg 5. The standard deviation in the weights of female brown bears is about 10 kg. Approximately what % of the female brown bears have weights that are within 1.5 standard deviation of the mean? = 88% mean 23% + 42% + 23% 6. The height of adult males are approximately normally distributed with mean 69.5 and standard deviation 2.5. What % of adult males are between 67 in. and 74.5 in. tall σ =2.5 34% 34% 2.35% 13.5% 62 64.5 67 69.5 13.5% 2.35% 72 74.5 77 34% + 34% + 13.5% = 81.5% 7. In a group of 2000 adult males, about how many would you expect to be taller than 6 ft? (or 72 in.) 34% 34% 2.35% 13.5% 62 64.5 67 69.5 13.5% 2.35% 72 74.5 77 2.35% + 13.5% = 15.85% (2000)(0.1585) = 317 8. The scores on the Algebra 2 final are approximately normally distributed with a mean of 150 and a standard deviation of 15. What % of the students who took the tests scored about 180? σ =15 34% 34% 2.35% 13.5% 105 120 135 150 13.5% 2.35% 165 180 195 = 2.35% 9. The scores on the Algebra 2 final are approximately normally distributed with a mean of 150 and a standard deviation of 15. If 250 students took the final, approximately how many scored above 135? 34% 34% 13.5% 2.35% 2.35% 13.5% 105 120 135 150 165 180 195 (34%+34%+13.5%+2.35%) (250)(0.8385) = 209